When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's…

When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's surface at height 'h' then relation between 'h' and 'R' is [R = radius of the earth]
  1. $\mathrm{h}=\mathrm{R}(\sqrt{3}-1)$
  2. $\mathrm{h}=\mathrm{R}$
  3. $\mathrm{h}=\mathrm{R}(\sqrt{2}-1)$
  4. $\mathrm{h}=2 \mathrm{R}$

Solution

$\frac{g}{3}=\frac{G M}{(R+h)^{2}} \quad g=\frac{G M}{R^{2}}$ $\therefore \frac{g}{3}(R+h)^{2}=g R^{2}$ $(R+h)^{2}=3 R^{2}$ $\mathrm{R}+\mathrm{h}=\sqrt{3} \mathrm{R} \quad \therefore \mathrm{h}=\sqrt{3} \mathrm{R}-\mathrm{R}=(\sqrt{3}-1) \mathrm{R}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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